If $$f$$ be a function given by $$f\left( x \right) = 2{x^2} + 3x - 5.$$     Then $$f'\left( 0 \right) = mf'\left( { - 1} \right),$$     where $$m$$ is equal to :

Solution :
We first find the derivatives of $$f\left( x \right)$$  at $$x = - 1$$   and at $$x = 0.$$
We have,
$$\eqalign{ & f'\left( { - 1} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( { - 1 + h} \right) - f\left( { - 1} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {2{{\left( { - 1 + h} \right)}^2} + 3\left( { - 1 + h} \right) - 5} \right] - \left[ {2{{\left( { - 1} \right)}^2} + 3\left( { - 1} \right) - 5} \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} - h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {2h - 1} \right) \cr & = 2\left( 0 \right) - 1 \cr & = - 1 \cr & {\text{and }}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {2{{\left( {0 + h} \right)}^2} + 3\left( {0 + h} \right) - 5} \right] - \left[ {2{{\left( 0 \right)}^2} + 3\left( 0 \right) - 5} \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} + 3h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {2h + 3} \right) \cr & = 2\left( 0 \right) + 3 \cr & = 3 \cr & {\text{Clearly, }}f'\left( 0 \right) = - 3f'\left( { - 1} \right) \cr} $$